# Does two quantum states interacting via some quantum interaction, always gets entangled?

We know that two quantum states, can not get entangled via local classical interactions/communication (LOCC). However, do two quantum states locally interacting via quantum interactions always get entangled?

Suppose, two Quantum systems, initially in the unentangled state, interact locally with a mediator, and no entanglement is generated between the two because of interaction, can we unambiguously determine that the mediator was classical in nature?

• what is a “mediator” in this case? Jun 30, 2022 at 11:44
• Say, the agent that locally interacts with the two systems and transmits information, like a field for example. Jun 30, 2022 at 12:44
• what does "locally interacting via quantum interactions" mean precisely here? Are you referring to an interaction via a specific type of Hamiltonian (eg one with pairwise interactions)?
– glS
Jul 2, 2022 at 8:12

# Answer

However, do two quantum states locally interacting via quantum interactions always get entangled?

No. I can give you an example using the $$CNOT$$ gate for qubits which is defined by the following relation:

$$CNOT |0\rangle |0\rangle = |0 \rangle |0\rangle$$ $$CNOT |0\rangle |1\rangle = |0 \rangle |1\rangle$$ $$CNOT |1\rangle |0\rangle = |1 \rangle |1\rangle$$ $$CNOT |1\rangle |1\rangle = |1 \rangle |0\rangle$$

i.e. the second qubit is flipped if the first one is in state $$|1\rangle$$. Clearly none of the above states is entangled after operating with CNOT.This is because the state can be written in the form $$|\cdot\rangle_{q_1}|\cdot\rangle_{q_2}$$ where $$q_1$$ and $$q_2$$ are qubits 1 and 2.

Now, look at the not-entangled state $$\frac{1}{\sqrt 2}(|0\rangle + |1\rangle)\otimes |1\rangle = \frac{1}{\sqrt 2} (|0\rangle|1\rangle + |1\rangle|1\rangle)$$

after acting with $$CNOT$$ we obtain: $$CNOT \frac{1}{\sqrt 2} (|0\rangle|1\rangle + |1\rangle|1\rangle) = \frac{1}{\sqrt 2} (|0\rangle|1\rangle + |1\rangle |0\rangle)$$

using the defining relations of the CNOT gate. The state $$\frac{1}{\sqrt 2} (|0\rangle|1\rangle + |1\rangle |0\rangle)$$ is called Bell state and it is a maximally entangled state. Hence, you can see that the final level of entanglement also depends on the initial state that experiences the interaction.

## Extension: Entangling Power

After doing some more reading, I came across the concept of entangling power. For a gate $$U$$, the entangling power $$K$$ is defined as [1]:

$$K_{E}(U)=\max _{|\phi\rangle,|\psi\rangle} E(U(|\phi\rangle|\psi\rangle))$$

This quantity is the maximum attainable entanglement $$E$$ maximised over all states $$|\phi\rangle|\psi\rangle$$.